Multiply the following rational expressions and simplify the result. $\dfrac{4}{k^2+14kn+49n^2} \cdot \dfrac{k^2-49n^2}{4k-28n}=$
Let's first factor the numerators and denominators of each expression separately. [Why are we doing this?] The numerator, $4$, of the first expression cannot be factored further. The denominator, $k^2+14kn+49n^2$, of the first expression can be factored as $(k+7n)(k+7n)$ using the perfect squares pattern. The numerator, $k^2-49n^2$, of the second expression can be factored as $(k+7n)(k-7n)$ by the difference of squares pattern. The denominator, $4k-28n$, of the second expression can be factored as $4(k-7n)$ by factoring out $4$. Now the product looks as follows: $\dfrac{4}{(k+7n)(k+7n)} \cdot \dfrac{(k+7n)(k-7n)}{4(k-7n)}$ To multiply two rational expressions, we multiply across, then simplify: [What's that?] $\phantom{=} \dfrac{4}{(k+7n)(k+7n)} \cdot \dfrac{(k+7n)(k-7n)}{4(k-7n)} $ $\begin{aligned} &= \dfrac{4 \cdot (k+7n)(k-7n)}{(k+7n)(k+7n) \cdot 4(k-7n)} &\text{Multiply across.}\\\\\\\\ &= \dfrac{{\cancel{4}} {\cancel{(k+7n)}}{\cancel{(k-7n)}}}{{\cancel{(k+7n)}}(k+7n) {\cancel{4}}{\cancel{(k-7n)}}} &\text{Cancel out common factors.}\\\\\\\\ &=\dfrac{1}{k+7n} \end{aligned}$ Therefore, the simplified form of the product is $\dfrac{1}{k+7n}$.